Passing arguments by value

This is a discussion on Passing arguments by value within the C++ Programming forums, part of the General Programming Boards category; Code: #include <iostream> using namespace std; int add (int i) { i=i+100; return i; } int main() { int n=1; ...

  1. #1
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    Passing arguments by value

    Code:
    #include <iostream>
    using namespace std;
    int add (int i)
    {
    	i=i+100;
    	return i;
    }
    int main()
    {
    int n=1;
    add(n);
    cout << "n:" << n << endl;
    n=add(n);
    cout << n << endl;
    	system("pause");
    	return 0;
    }
    Why is result 1 in statement add(n) and why it is 101 in n=add(n).I have also noticed that if I write
    Code:
    cout << "n:" << add(n) << endl;
    instead
    Code:
     add(n);
    result is 101?

    In which cases it will return 1 and why?

  2. #2
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    Since the parameter is passed by value, the function cannot modify it. The function modifies a copy, then returns the result. So do this instead:
    Code:
    int n=1;
    n = add(n); // Assign n to what the function returns
    bit∙hub [bit-huhb] n. A source and destination for information.

  3. #3
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    Quote Originally Posted by bithub View Post
    Since the parameter is passed by value, the function cannot modify it. The function modifies a copy, then returns the result. So do this instead:
    Code:
    int n=1;
    n = add(n); // Assign n to what the function returns
    But why I need to assign it to n?

  4. #4
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    Because the add function isn't modifying the passed value. If you want the add function to modify the value that is passed in, then do this:
    Code:
    int add (int& i)
    {
    	i=i+100;
    	return i;
    }
    This way i is passed by reference instead of by value. Now any changes add makes will change the caller's variable.
    bit∙hub [bit-huhb] n. A source and destination for information.

  5. #5
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    Code:
    int n = 1;
    add(n);
    In the case where the variable is passed by value, the above code will change the local version of the variable n (called i within the function itself) however, since this new value (although returned to the caller) is not assigned back to n in main, it (n) maintains its old value of 1.

    Code:
    int n = 1;
    n=add(n);
    In this case, the returned value 101 is assigned back to n overwritting what was there (a 1). Thus n will now contain 101 after the function call.

    Code:
    int n = 1;
    cout << "n:" << add(n) << endl;
    This will print the value returned by the function (101) but will not update n, n will still remain 1 after the call to add because it hasn't been changed.
    "Owners of dogs will have noticed that, if you provide them with food and water and shelter and affection, they will think you are god. Whereas owners of cats are compelled to realize that, if you provide them with food and water and shelter and affection, they draw the conclusion that they are gods."
    -Christopher Hitchens

  6. #6
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    I understand it now.

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